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Source: Slovenia TST 2005 Test 1 Problem 2

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#1 h Apr 5, 2021, 2:35 AM • 2 Y

Y by FaThEr-SqUiRrEl, megarnie

Find all functions $f:\mathbb R^+\to\mathbb R^+$ such that for any $x,y>0$ ,
$x^2\left(f(x)+f(y)\right)=(x+y)f\left(f(x)y\right).$

This post has been edited 1 time. Last edited by jasperE3, Apr 5, 2021, 2:36 AM

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pco

#2 h Apr 5, 2021, 6:19 AM • 1 Y

Y by FaThEr-SqUiRrEl

jasperE3 wrote:

Find all functions $f:\mathbb R^+\to\mathbb R^+$ such that for any $x,y>0$ ,
$x^2\left(f(x)+f(y)\right)=(x+y)f\left(f(x)y\right).$

Let $P(x,y)$ be the assertion $x^2(f(x)+f(y))=(x+y)f(f(x)y)$

If $\exists u>0$ such that $f(u)=u$ , then :
$P(u,1)$ $\implies$ $u^2+u(f(1)-1)-1=0$ and so at most one such positive $u$

$P(x,x)$ $\implies$ $f(xf(x))=xf(x)\quad\forall x>0$ and so $f(x)=\frac ux$ $\forall x>0$

Plugging this back in original equation, we get $u=1$ and so $\boxed{f(x)=\frac 1x\quad\forall x>0}$ , which indeed fits.

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Jjesus

#3 h Apr 5, 2021, 11:50 PM • 1 Y

Y by FaThEr-SqUiRrEl

pco wrote:

jasperE3 wrote:

$P(x,x)$ $\implies$ $f(xf(x))=xf(x)\quad\forall x>0$ and so $f(x)=\frac ux$ $\forall x>0$

Why?

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#4 h Apr 6, 2021, 12:22 AM • 2 Y

Y by pco, FaThEr-SqUiRrEl

We have that $xf(x)$ must be constant, since there is at most one fixed point of $f$ and $xf(x)$ is a fixed point, thus $xf(x)=u\Rightarrow f(x)=\frac ux$ .

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#5 h Apr 6, 2021, 12:40 PM • 1 Y

Y by FaThEr-SqUiRrEl

Denoting the given assertion of $P(x,y)$ ; suppose for some $x_1,x_2$ ; $f(x_1)=f(x_2)$ . Inspecting $P(x_1,1)$ and $P(x_2,1)$ , we find that
$\frac{x_1^2}{x_1+1} = \frac{x_2^2}{x_2+1}\implies x_1^2x_1+x_1^2 = x_1x_2^2+x_2^2 \Leftrightarrow \left(x_1-x_2\right)\left(x_1+x_2+x_1x_2\right)=0.$ Since $x_1,x_2>0$ , it follows $x_1=x_2$ . That is, $f$ is injective. Now, inspecting $P(1,1)$ , we obtain $f(1)=f(f(1))$ , which together with injectivity yields $f(1)=1$ . Finally, $P(1,y)$ gives $f(y) =1/y$ for every $y>0$ .

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#6 h Jun 8, 2022, 12:18 PM • 2 Y

Y by Mango247, Mango247

Let $P(x,y)$ denote the given assertion.

$P(x,x): 2x^2f(x)=2xf(xf(x))\implies f(xf(x))=xf(x)$ .

Let $f(k)=k$ for some $k>0$ .

$P(k,1): k^2(k+f(1))=(k+1)f(f(k))=k(k+1)$ .

Dividing both sides by $k$ gives $k(k+f(1))=k+1\implies k^2+f(1)k=k+1$ , so $k$ must be a root of $x^2+(f(1)-1)x-1=0$ . This has at most $1$ positive root.

So there is at most $1$ value of $k$ .

This implies $xf(x)=k$ for each $x$ , so $f(x)=\frac{k}{x}$ .

So $x^2k\frac{x+y}{xy}=(x+y)f\left(\frac{ky}{x}\right)=(x+y)\frac{x}{y}$ , so $\frac{x^2k}{xy}=\frac{x}{y}\implies k=1$ .

Thus, the only solution is $\boxed{f(x)=\frac{1}{x}}$ , which works.

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#7 h Jun 8, 2022, 12:40 PM • 1 Y

Y by pco

Let $P(x,y)$ be assertion.
$P(1,1) : 2f(1) = 2f(f(1))$
$P(f(1),1) : {f(1)}^2 * 2f(1) = (f(1) + 1)f(1) \rightarrow f(1) = 1$
$P(1,y) : 1 + f(y) = (1 + y)f(y) \rightarrow f(y) = \frac{1}{y}$

This post has been edited 2 times. Last edited by Darkztar, Jun 8, 2022, 12:41 PM

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ZETA_in_olympiad

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ZETA_in_olympiad

#8 h Jun 12, 2022, 4:31 PM • 3 Y

Y by Mango247, Mango247, Mango247

Let $P(x,y)$ denote the assertion.

$P(1,f(1)^{-1})$ gives $f(f(1)^{-1})=1.$ Then $P(f(1)^{-1},1)$ forces $f(1)=1.$ Now $P(1,x)$ implies $f\equiv \text{Id}^{-1}$ which works.

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#9 h Apr 27, 2025, 8:23 PM

Y by

ZETA_in_olympiad wrote:

Let $P(x,y)$ denote the assertion.

$P(1,f(1)^{-1})$ gives $f(f(1)^{-1})=1.$ Then $P(f(1)^{-1},1)$ forces $f(1)=1.$ Now $P(1,x)$ implies $f\equiv \text{Id}^{-1}$ which works.

didn't prove bijectivity at $1$

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#10 h Apr 27, 2025, 8:27 PM

Y by

jasperE3 wrote:

ZETA_in_olympiad wrote:

Let $P(x,y)$ denote the assertion.

$P(1,f(1)^{-1})$ gives $f(f(1)^{-1})=1.$ Then $P(f(1)^{-1},1)$ forces $f(1)=1.$ Now $P(1,x)$ implies $f\equiv \text{Id}^{-1}$ which works.

didn't prove bijectivity at $1$

i think $f(1)^{-1}$ means $\frac{1}{f(1)}$ , not $f$ inverse of $1$

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#11 h Apr 27, 2025, 8:34 PM

Y by

megarnie wrote:

jasperE3 wrote:

ZETA_in_olympiad wrote:

Let $P(x,y)$ denote the assertion.

$P(1,f(1)^{-1})$ gives $f(f(1)^{-1})=1.$ Then $P(f(1)^{-1},1)$ forces $f(1)=1.$ Now $P(1,x)$ implies $f\equiv \text{Id}^{-1}$ which works.

didn't prove bijectivity at $1$

i think $f(1)^{-1}$ means $\frac{1}{f(1)}$ , not $f$ inverse of $1$

true thank you

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